Problem 52 - Algebra A Combined Function

Question

The graph of each equation is an ellipse. Determine which distance is longer, the distance between the \(x\) -intercepts or the distance between the y-intercepts. How much longer? $$ x^{2}+4 y^{2}=36 $$

Step-by-Step Solution

Answer

The x-intercept distance is 6 units longer.

Step 1: Identify the Form of the Equation

The given equation is \(x^{2} + 4y^{2} = 36\). This is an equation of an ellipse. We need to rewrite it in standard form to identify the axes lengths.

Step 2: Rewrite in Standard Form

Divide every term in the equation by 36 to get it into the standard form \(\frac{x^2}{36} + \frac{y^2}{9} = 1\). The semi-major axis is along the x-axis.

Step 3: Determine Length of the x-intercepts

For the x-intercepts, let \(y = 0\): \(\frac{x^2}{36} = 1\) leads to \(x^2 = 36\), so \(x = \pm 6\). Thus, the distance between the x-intercepts is \(6 - (-6) = 12\).

Step 4: Determine Length of the y-intercepts

For the y-intercepts, let \(x = 0\): \(\frac{y^2}{9} = 1\) leads to \(y^2 = 9\), so \(y = \pm 3\). Thus, the distance between the y-intercepts is \(3 - (-3) = 6\).

Step 5: Comparison and Calculation of Difference

Calculate the difference between the x-intercept distance and the y-intercept distance: \(12 - 6 = 6\). Therefore, the distance between the x-intercepts is longer by 6 units.

Key Concepts

x-intercepts

In mathematics, finding the x-intercepts of an ellipse involves determining where the graph cuts through the x-axis. Simply put, these are the points where the value of y equals zero.
For an ellipse described by its standard form equation like \( \frac{x^2}{36} + \frac{y^2}{9} = 1 \), the x-intercepts can be found by setting y to zero. This simplifies the equation to \( \frac{x^2}{36} = 1 \).
By solving the simplified equation, you'll get \( x^2 = 36 \), which leads to \( x = \pm 6 \).
The x-intercepts for this particular ellipse are therefore at (-6, 0) and (6, 0). The distance between these points, often referred to as the 'span' along the x-axis, is calculated by simply finding the difference: \( 6 - (-6) = 12 \).

• x-intercepts occur where y=0.
• Calculated by: \( x^2 = (semi-major\ axis)^2 \).
• In this example, intercepts are -6 and 6, span is 12 units.

y-intercepts

Similarly, y-intercepts for an ellipse occur where the x-axis is crossed, meaning the value of x is zero.
Consider the standard form equation \( \frac{x^2}{36} + \frac{y^2}{9} = 1 \), which can help us find y-intercepts by substituting x with zero.
With this substitution, the equation becomes \( \frac{y^2}{9} = 1 \).
Solve this to find \( y^2 = 9 \), leading to \( y = \pm 3 \). The y-intercepts for this ellipse are therefore at (0, -3) and (0, 3).
The span between these intercepts is calculated similarly to x-intercepts: \( 3 - (-3) = 6 \) units.

• Y-intercepts occur where x=0.
• Calculated by: \( y^2 = (semi-minor\ axis)^2 \).
• In this example, intercepts are -3 and 3, span is 6 units.

standard form of ellipse

The standard form of an ellipse is crucial in identifying its geometric properties. The general equation takes the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) represent the lengths of the semi-major and semi-minor axes, respectively.
For the equation \( x^2 + 4y^2 = 36 \), starting by dividing each term by 36 yields \( \frac{x^2}{36} + \frac{y^2}{9} = 1 \), which is the standard form.
This transformed equation allows for easier identification of the ellipse's axes:

• \( a^2 = 36 \Rightarrow a = 6 \), the semi-major axis, representing the larger span along the x-axis.
• \( b^2 = 9 \Rightarrow b = 3 \), the semi-minor axis, representing the smaller span along the y-axis.

This form is incredibly useful for visualizing and calculating the intercepts and the overall shape of the ellipse.

semi-major axis

The semi-major axis is a key feature of an ellipse, dictating its longest diameter. It runs from the center to the farthest point on the ellipse, parallel to the x-axis in our example.
Using the standard form equation \( \frac{x^2}{36} + \frac{y^2}{9} = 1 \), the semi-major axis can be understood as the \( a \) value. Here, \( a^2 = 36 \), leading us to \( a = 6 \).
This axis is crucial because it directly affects the distance between the x-intercepts, which we calculated to be 12 units.

• Larger diameter of the ellipse.
• Inferenced by highest denominator in standard form.
• Critical in calculating x-intercept span.

The semi-major axis, therefore, not only defines the ellipse's size but also gives insight into its orientation relative to the coordinate axes.